Question

rationalise the denominater 3 √8 -5√6 by 2√8+ 3√6​

Answer 1

Given:

The expression -

$\bf \dashrightarrow \: \dfrac{ 3\sqrt{8}-5\sqrt{6} }{ 2\sqrt{8} + 3\sqrt{6} }$

What To Find:

We have to -

• Rationalise the denominator.

Solution:

• Finding the rationalising factor of the denominator.

Here the rationalising factor is -

$\sf \implies 2\sqrt{8} - 3\sqrt{6}$

• Solving the expression by multiplying with the rationalising factor.

Multiply the expression,

$\sf \implies \dfrac{ 3\sqrt{8} - 5\sqrt{6} }{ 2\sqrt{8} + 3\sqrt{6} } \times \dfrac{ 2\sqrt{8} - 3\sqrt{6} }{ 2\sqrt{8} - 3\sqrt{6} }$

Take them as common,

$\sf \implies \dfrac{ 3\sqrt{8} - 5\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6} }{ 2\sqrt{8} + 3\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6}}$

• Solving the numerator.

$\sf \implies \dfrac{ 3\sqrt{8}( 2\sqrt{8} - 3\sqrt{6} ) - 5\sqrt{6} ( 2\sqrt{8} - 3\sqrt{6} ) }{ 2\sqrt{8} + 3\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6}}$

Solving the first brackets,

$\sf \implies \dfrac{ 48 - 9\sqrt{48} - 5\sqrt{6} ( 2\sqrt{8} - 3\sqrt{6} ) }{ 2\sqrt{8} + 3\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6}}$

Solving the second brackets,

$\sf \implies \dfrac{ 48 - 9\sqrt{48} - 10\sqrt{48} + 90 }{ 2\sqrt{8} + 3\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6}}$

Rearrange the terms,

$\sf \implies \dfrac{ 48 + 90 - 9\sqrt{48} - 10\sqrt{48}}{ 2\sqrt{8} + 3\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6}}$

Solve the like terms,

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ 2\sqrt{8} + 3\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6}}$

• Solving the denominator.

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ 2\sqrt{8} + 3\sqrt{6} \times 2\sqrt{8} - 3\sqrt{6}}$

Using the identity - (a - b) (a + b) = a² - b²

Where -

• a = 2√8
• b = 3√6

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ (2\sqrt{8})^2 - (3\sqrt{6})^2}$

Solve the first brackets,

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ (2 \times 2 \times \sqrt{8} \times \sqrt{8}) - (3\sqrt{6})^2}$

Multiply the numbers in the first brackets,

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ 32 - (3\sqrt{6})^2}$

Solve the second brackets,

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ 32 - (3 \times 3 \times \sqrt{6} \times \sqrt{6})}$

Multiply the numbers in the second brackets,

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ 32 - 54}$

Subtract the numbers,

$\sf \implies \dfrac{ 138 - 19\sqrt{48}}{ - 22}$

Also written as,

$\sf \implies - \dfrac{ 138 - 19\sqrt{48}}{ 22}$

Final Answer:

∴ Hence, the answer is $\sf - \dfrac{ 138 - 19\sqrt{48}}{ 22}$.